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Chi-Square Test of Independence

The Chi-Square Test of Independence Calculator helps you determine whether there is a significant relationship between two categorical variables. It analyzes whether the observed frequency distribution differs significantly from the expected distribution, assuming the variables are independent. This test is widely used in fields like social sciences, market research, and healthcare to analyze survey data, clinical outcomes, and demographic relationships. Common applications include examining relationships between demographic factors and preferences, testing associations between treatments and outcomes, or analyzing connections between categorical variables in survey responses. Click here to populate the sample data for a quick example.

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Chi-Square Test of Independence

Definition

Chi-Square Test of Independence examines whether there is a significant association between two categorical variables. It tests whether the observed frequencies in a contingency table differ significantly from the frequencies we would expect if there were no relationship between the variables.

Formula

Test Statistic:

χ2=∑(Oij−Eij)2Eij\chi^2 = \sum\frac{(O_{ij} - E_{ij})^2}{E_{ij}}

Where:

  • OijO_{ij} = observed frequency in cell (ii,jj)
  • EijE_{ij} = expected frequency in cell (ii,jj)
  • Eij=(row i total)(column j total)grand totalE_{ij} = \frac{(\text{row }i\text{ total})(\text{column }j\text{ total})}{\text{grand total}}

Modified Formula with Yates Continuity Correction (for 2x2 tables):

χ2=∑(∣Oij−Eij∣−0.5)2Eij\chi^2 = \sum\frac{(|O_{ij} - E_{ij}| - 0.5)^2}{E_{ij}}

Key Assumptions

Random Sampling: Data must be randomly sampled
Independence: Observations must be independent
Sample Size: Expected frequenciesshould be ≥5\geq 5
Mutual Exclusivity: Categories must be mutually exclusive

Practical Example

Step 1: State the Data

Contingency table of Gender and Product Preference:

LikeDislikeTotal
Male403070
Female305080
Total7080150
Step 2: State Hypotheses
  • H0H_0: Gender and Preference are independent
  • HaH_a: Gender and Preference are not independent
  • α=0.05\alpha = 0.05
Step 3: Calculate Expected Frequencies
  • Male, Like: E11=70×70150=32.67E_{11} = \frac{70 \times 70}{150} = 32.67
  • Male, Dislike: E12=70×80150=37.33E_{12} = \frac{70 \times 80}{150} = 37.33
  • Female, Like: E21=80×70150=37.33E_{21} = \frac{80 \times 70}{150} = 37.33
  • Female, Dislike: E22=80×80150=42.67E_{22} = \frac{80 \times 80}{150} = 42.67
Step 4: Calculate Chi-Square Statistic
χ2=(∣40−32.67∣−0.5)232.67+(∣30−37.33∣−0.5)237.33+(∣30−37.33∣−0.5)237.33+(∣50−42.67∣−0.5)242.67=5.02\chi^2 = \frac{(|40-32.67|-0.5)^2}{32.67} + \frac{(|30-37.33|-0.5)^2}{37.33} + \frac{(|30-37.33|-0.5)^2}{37.33} + \frac{(|50-42.67|-0.5)^2}{42.67} = 5.02
Step 5: Calculate Degrees of Freedom

df=(r−1)(c−1)=(2−1)(2−1)=1df = (r-1)(c-1) = (2-1)(2-1) = 1

Step 6: Draw Conclusion

At α=0.05\alpha = 0.05 with df=1df = 1, the critical value is 3.8413.841. Since χ2=5.02>3.841\chi^2 = 5.02 \gt 3.841, we reject H0H_0. There is sufficient evidence to conclude that Gender and Product Preference are not independent (pp-value =0.008= 0.008).

Effect Size

Cramer's V measures the strength of association:

V=χ2n(min⁡(r,c)−1)V = \sqrt{\frac{\chi^2}{n(\min(r,c)-1)}}

For our example:

V=5.02150(1)=0.18V = \sqrt{\frac{5.02}{150(1)}} = 0.18

For 2×22\times 2 tables:

  • Small effect: V≈0.10V \approx 0.10
  • Medium effect: V≈0.30V \approx 0.30
  • Large effect: V≈0.50V \approx 0.50

With V=0.18V = 0.18, this indicates a small effect size, suggesting a weak association between Gender and Product Preference in our sample.

Verification

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